CHEMISTRY COMMUNITY

megansardina2G

The half life equation for a second order reaction is: t(1/2) = 1/(K)([A]o) This can be derived from some of the equations that we got in class: t=1/K (1/[A] - 1/[A]o) to get K= 1/(t(1/2))([A]o) and then switch the K and t(1/2) you can also solve for the time it takes for decay that isn't to the hal...

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They are not part of the overall rate law because they are produced in one step and consumed in another, later step.

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Similar to when the amount of product produced was based on the limiting reactant, the rate of a reaction is dependent on the slowest step. It doesn't matter how much you change the faster reaction, the slower reaction is still going to be in control because you want to take into account the entire ...

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No matter what situation, I think the easiest way to determine the units is by looking at the concentrations, since the rate units will always be mol • L-1 • s-1, so whatever leads to that answer multiplied by the concentration's units ends up being k.

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Reaction: H+ (aq) + OH- (aq) --> H2O (l) Split into reduction and oxidation reactions: H+ is gaining an electron so it is being reduced, OH- is losing an electron so it is being oxidized --> remember OILRIG (oxidation is losing, reduction is gaining) Reduction: 4 H+ + O2 + 4e- --> 2 H2O Oxidation: 4...

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Why are products never included in rate laws?

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Pure water is always neutral, since the [H+] and [OH-] concentrations will be equal. If you did that calculation method to get to the pH and it was less than 7, then a pH of 7 would be considered basic. If you got a pH value of greater than 7, then a pH of 7 would be considered acidic. I believe tha...

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Q has no units (like K) so when you are multiplying and dividing these values, the units end up cancelling out.

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Disproportionation is a chemical reaction where a molecule is transformed into two or more dissimilar products. You can tell if a molecule will disproportionate by looking at the ∆Gº of the reaction. If ∆Gº is positive, then the reaction is not spontaneous, K<1, and reactant formation is favored. If...

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At constant temperature, ∆H=0, as there is no change, so I'm not sure why you would be asked to calculate ∆H at constant temperature. Under only constant pressure, ∆H = q (you can solve for ∆H using formulas for/with q).

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The first two are the same equation. They are used under constant pressure. (for the second: pressure is constant, so it can be taken outside the integral. The integral of dV is V, and V2-V1 is ∆V, thus the first equation: w=-p∆V) The third equation is used when the reaction is isothermal, but press...

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You can balance the two skeletal equations using H2O, since they are aqueous solutions. The first equation is O3 --> O2 Adding H2O and OH- will balance the oxygens and hydrogens on either side, and then you balance the charge to result in: H2O + O3 + 2e- --> O2 + 2OH- The second equation is Br- --> ...

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Why does the entropy of the universe always increase? Is there some sort of justification? I know that entropy of a system can decrease, but entropy of the universe still increases.

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How are reactions that have a positive ∆G carried out, as they are not spontaneous? (Especially in biological systems)

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What is a good definition for “free energy”? I’m a bit confused on what the problems are talking about when the energy is “free”.

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Yes, it is important to understand all of the parts of the phase change graph, as when you are changing between multiple phases you have to use each step. For example, if you want to go from solid to gas, you must calculate the total delta H in five steps. the mC(delta T) with the solid specific hea...

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Degeneracy is the number of ways of achieving a given energy state. The relationship between degeneracy, W, and entropy, S, is the Boltzmann equation (S = Kb lnW)

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Kb is Boltzmann's constant. It is provided on the formula sheet.

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There are two parts to the question. Part (a) refers to the constant volume, and part (b) refers to the constant pressure. This results in different changes in temperature, final temperature, and difference changes in internal energy.

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Hi! So looking at the four graphs, you can solve this mainly by process of elimination. You know that the change in enthalpy for fusion is 10.0 kJ/mol and the change in enthalpy for vaporization is 20.0 kJ/mol. Since the heating rate is constant, the time for fusion (melting) would be twice as fast ...

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Since it was a part of the syllabus and included in the parts of the textbook to look over/read/do hw problems from, I would assume that we need to at least be familiar with it, although I can't say for sure. Better safe than sorry though!

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The solution/solutions manual either added the 10-3 onto the 3.1 or the problem was written so that it accidentally didn't have the 10-3. If it was supposed to be 3.1 and not 3.1x10^3 (as it is in the solutions manual), then the answer should have been 3.2x10^-15. (Basically, something was written i...

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The easiest and most important thing to remember is that state properties are independent of the path taken (to establish property or value). They depend only on the final and initial values, which are all you need to solve. This is contrasted by path functions (like work and heat), which are depend...

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State properties are quantities that are independent of how a substance was prepared. Work is dependent on path, so it is not a state function. Since work is not a state property/function, heat cannot be either, since the first law of thermodynamics states that the change in the initial energy of a ...

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Water is a liquid, which is a pure substance (along with solids); you cannot increase or decrease its concentration. Therefore, liquids are omitted when dealing with equilibrium constants.

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Hi! So for part (b), you have to solve for the new concentration in the increased volume flask. c1v1 =c2v2 will help you with this. c1= 0.0250 M, v1=200 mL, c2=?, v2=250 mL. So, c2 = (0.025 x 200)/(250) = 0.020 M. Since the acid is HCl, which is a strong acid, it is completely deprotonated and the m...

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K is temperature dependent, so a change in temperature will result in a change in equilibrium (that has to be restored, resulting in a shift in either the forward or backwards direction). For an endothermic reaction, Kc increases with an increase in temperature (and decreases with a decrease in temp...

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Yes. K does not change when the pressure changes, and since K is temperature dependent you know that the temperature is also constant. Therefore, the volume/concentration has to be the changing factor.

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Yes, because when the temperature changes the reaction shifts one way or another (endothermic reactions shift towards product formation, exothermic reactions shift toward reaction formation). K is temperature dependent and also changes to reestablish equilibrium, and the concentrations of the reacta...

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A quick way of doing this is determining whether the reaction is endothermic or exothermic. An endothermic reaction favors the formation of products, and an exothermic reaction favors the formation of reactants. Since delta H (enthalpy) is positive for (a), it is endothermic and therefore shifts the...

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Simply speaking, an oxoacid is an acid that contains oxygen. More specifically, it must contain an oxygen, hydrogen, and at least one other element; also, the oxygen has to be bonded to a hydrogen so that it can donate the H+ ion.

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When identifying the stronger acid, there are a few things to keep in mind. If there is no oxygen atom, then you judge the non-hydrogen atoms by the weaker bond length. If the central atom is the same and the oxygen atom number is different, the molecule with the higher oxygen atom number will be st...

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If a molecule is polar, it is able to form dipole-dipole intermolecular forces (and possibly hydrogen bonds) in addition to London Dispersion forces. With more forces and bonds, a molecule becomes harder to break, thus causing a higher boiling point to break the bonds and denature the molecule.

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I know that, when writing out the compound name, you write the ligands in alphabetical order, so I would stick with that same order for the compound formula. However, I do not think it is as important as writing out the ligands in the name.

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Since iodine is much larger than fluorine and has many more electrons, the London dispersion forces will be much stronger in CHI3 than CHF3. Therefore, CHI3 will have the higher melting point.

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Sigma bonds are always single bonds, or the first bond formed. In bonding pairs that share only a single bond, it will always be a sigma bond. However, if there is a double bond, there will be one sigma bond and one pi bond. Similarly, if there is a triple bond, there will be one sigma bond and two ...

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Hi! A refers to the central atom of the molecule, X refers to the atoms attached to the central atom in the molecule, and E refers to the lone pairs on the central atom. For example, XeF4 has four bonding pairs and two lone pairs, so it would be AX4E2. Hope this helps!

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Bond angles shouldn't be affected by differing bond strength. For example, a double bond between two atoms is stronger than a single bond, but this won't matter in regards to bond angles and shape. Hope this helps!

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The electron configuration for Ni is [Ar]3d^8 4s^2. Focusing on the 3d^8, since the d orbital is able to hold 10 electrons, due to Hund's rule, each electron suborbital has to be filled with one electron before they can double up with opposite spins. Therefore, there are 2 electrons that are unpaire...

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Hi! The difference between cis-dichloroethene and trans-dichloroethene are the sides on which the chlorine atom is on. In cis-dichloroethene, both chlorine atoms are on the right side of the double bonded carbon atoms, whereas in trans-dichloroethene, one chlorine atom is on the left and one is on t...

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You look at the differences in electronegativity. In the sixth edition textbook, figure 3.12 shows most of the values you'll need (and all for this problem). Even if you aren't looking at the exact values, you can estimate it using electronegativity trends. Electronegativity increases across a perio...

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Hi! As you stated, the ground state electron configuration of Cd is [Kr] 4d^10 5s^2. If you want the most likely ion for Cd, you need to take into consideration the electron configurations that are the most stable. The configurations are most stable when the orbitals are either half full or full, so...

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Polarizability increases with the increase in number of electrons. This is due to the decreased control that the nuclear charge has on the overall charge distribution in the atom. In small atoms, the electron cloud is small and dense, and the electrons have a stronger interaction/attraction with the...

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Neutrons do diffract because they have wavelength properties (like electrons and other particles), as stated in de Broglie's equation.

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Hi! So most of the time, the ideal lewis structure forms when all atoms in the molecule have a formal charge of 0. If this is not possible, then the formal charge should be the lowest possible option (closest to net of 0). When there needs to be a charge, the negative charge should be placed on the ...

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All the electrons across a row, or period, are added to the same shell. Protons generally increase at 1:1 ratio with electrons, resulting in a increasing attraction between the electron shells and the nucleus (since the effect of the increasing proton number is not balanced with the effect of the si...

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Covalent bonds (both polar and nonpolar) are stronger than ionic bonds. Unlike the unequal "stealing" of electrons that occurs during ionic bond interactions, covalent bonds share electrons, which provides a more stable molecule with lower potential energy, thus making the bond stronger an...

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The principle quantum number (n) usually determines the order of electron configuration listings, so 4f would be before 5d and 5d would be before 6s. The reason why 5p and 5s are not listed are that they are included in the electron configuration for Xe, so the electron configuration for tungsten wo...

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Ionization energy can be defined as the amount of energy required to remove one (or more) electron(s) from a neutral atom to form a positively charged ion. This increases across a period due to the increasing number of valence electrons across a period. For example, an alkali metal such as potassium...

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A nodal plane is an area with zero probability of finding an electron. You should know that the s orbital has 0 nodal planes, the p orbital has 1, the d orbital has 2, and the f orbital has 3. (for planes beyond f, the nodal plane number can be found by n-1).

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Hi! The d-orbitals aren't actually labeled with p's like px, py, and pz, since that notation only applies to the p orbitals. (I believe s is different as well, not px) Rather, the five d-orbitals follow this organization: -3 have 4 lobes of electrons located in the xy-, yz-, and zx- planes (dxy, dyz...

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No, I do not think that we have to read every section in the textbook. Test two covers all the quantum material that we have previously covered in class (as of 10-19-18), which is up to quantum numbers. If we have not covered a part of the material in class, even if it is in the textbook, we do not ...

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HI! So this problem can be solved while referring to section 2.6 of the sixth edition textbook. In (a), the electron configuration represents an excited state. A carbon atom would fill the 1s and 2s orbitals, both with notation ↑↓, and have two electrons in the 2p orbital. This orbital is the focus ...

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"Formula units" is simply a term, similar to molecules or atoms. It references to the ions in an ionic compound, whereas "molecule" refers to two or more elements that are covalently bonded. Whether the problem asks for formula units, molecules, or atoms, you will multiply the nu...

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I don't think you're missing a specific reason other than frequency, by definition, cannot be negative, since frequency is the number of waves per second. The purpose of the negative sign in this example isn't really about numerical values, but to distinguish between gaining energy (+) and losing en...

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Hi! So for problem 15, you have to solve for the frequency of the line before you use the Rydberg equation to solve for n2. You already know n1=1, since the line is in the ultraviolet spectrum, and therefore a part of the Lyman series, where n1 always equals 1. (If you want to fact-check this, it ca...

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Hi! I took the test on Tuesday (Discussion 3A) and I think that if you worked on the assigned homework problems and successfully completed the post assessment questions on the modules you will be prepared! That is mainly what I did. If you want a little bit more review, I suggest looking over the ex...

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3 posts per week, for a total of 30 points for 10 weeks. It is posted on the grading section of his syllabus, on the second page.

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I think that, when discussing calculations, the difference between atoms, molecules, and formula units is not important. In any problem that asks you to solve for a certain amount of any of these, you would multiply the number of moles by Avogadro's constant, 6.022 x 10^23. To break down the differe...

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For this problem, you are looking for the molar mass of the sulfide of the metal in the compound M(OH)2. The M is unknown, so the first step is to solve for M, or find the metal in the compound. Since the molar mass of this compound is given, you can deduce the metal's molar mass by subtracting the ...

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Hi! The third equation in L35 in my textbook was given as: Fe3Br8 + Na2CO3 -----> NaBr + CO2 + Fe3O4 (I have the sixth edition textbook, but I think the equation should be the same for this problem, even if you have the seventh) The equation that you wrote down in your question had FeBr2 in place of...

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